What is the four-momentum of a photon?

[jbergman]

TL;DR Summary

Does a single photon still exhibit a doppler shift?

I'm trying to understand the Energy-Momentum relativistic relationships for a light particles. It is commonly said that the Energy of a photon depends on the observer by the relationship $E = - \mathbf{p} \cdot \mathbf{u}$ where p is the 4-momentum of the source emitting light particles and u is the 4-velocity of the observer.

The way I understand this physical effect to occur is via relativistic doppler shift that results in a change in frequency as described here, https://en.wikipedia.org/wiki/Relativistic_Doppler_effect

However, if you have a single photon emitted by a source do these equations still hold? I am struggling to see how the reasoning would still apply since the doppler effect relies the distance between multiple wavefronts from different photons, I assume.

 

[PeroK]

Summary:: Does a single photon still exhibit a doppler shift?

I'm trying to understand the Energy-Momentum relativistic relationships for a light particles. It is commonly said that the Energy of a photon depends on the observer by the relationship $E = - \mathbf{p} \cdot \mathbf{u}$ where p is the 4-momentum of the source emitting light particles and u is the 4-velocity of the observer.

The way I understand this physical effect to occur is via relativistic doppler shift that results in a change in frequency as described here, https://en.wikipedia.org/wiki/Relativistic_Doppler_effect

However, if you have a single photon emitted by a source do these equations still hold? I am struggling to see how the reasoning would still apply since the doppler effect relies the distance between multiple wavefronts from different photons, I assume.

You're mixing your theories here. In classical EM, light is a wave, with a wavelength given by the distance between wavefronts. In SR, photons are modeled as massless particles, with the frequency and wavelength being parameters related to the energy-momentum of the photon. The wavelength is in no way the distance between adjacent photons. The number of photons per second is the intensity of the light.

 

[jbergman]

Two responses.
1. So you are saying that a single photon would have different energies as viewed by different observers due to the energy-momentum relations and that because of the energy frequency relationship it would be doppler shifted? Or do single photons not have associated frequencies?

2. I find falling back to the Energy-Momentum relationships unsatisfying. My understanding based on the wikipedia article on the Mass-Energy equivalence, that this was derived by looking at doppler shifts associated with photon emissions. The article on the relativistic doppler effect mentioned above also talks about the distance between wavefronts and time dilation. If you are going to fallback onto the energy momentum relationships, can you tell me how you would derive them or point me to a source that explains this?

 

[Vanadium 50]

So you are saying that a single photon would have different energies as viewed by different observers

Why not? A single automobile has different energies as viewed by different observers. This has nothing to do with relativity.

 

[jbergman]

Why not? A single automobile has different energies as viewed by different observers. This has nothing to do with relativity.

For one we are dealing with a massless particle whose speed is the same in all reference frames. I am trying to wrap my head around how to think about the energy of light particles other than asserting that the time component of the 4-momentum is its energy.

 

[lomidrevo]

2. I find falling back to the Energy-Momentum relationships unsatisfying.

Bad for you, because energy and momentum themselves are not invariant. Only combined together into four-momentum can be properly transformed between frames.

1. So you are saying that a single photon would have different energies as viewed by different observers

Correct, energy is frame dependent, as well as momentum.

 

[Dale]

you are saying that a single photon would have different energies as viewed by different observers

Yes. At least I say that, @PeroK can speak for himself but the physics is pretty clear on this.

As you move, the frequency of spectral lines emitted as single photons from specific atomic transitions does indeed Doppler shift. So the idea is both theoretically justified and experimentally confirmed.

If you are going to fallback onto the energy momentum relationships, can you tell me how you would derive them or point me to a source that explains this?

My favorite way is to start with the four momentum $(E/c,\vec p)$ and then define the mass as the invariant norm $m^2 c^2=E^2/c^2-\vec p^2$. So the energy momentum relationship is easy. What is harder is motivating calling the thing on the left “mass” and showing that the four momentum is a legitimate four vector.

 

[PeroK]

For one we are dealing with a massless particle whose speed is the same in all reference frames. I am trying to wrap my head around how to think about the energy of light particles other than asserting that the time component of the 4-momentum is its energy.

Isn't that just a prejudice of having learned classical physics first? Take the fundamental relation for any particle: $$E^2 = p^2c^2 + m^2c^4$$ Applied to a massless particle gives $$E = pc$$ Applied to a massive particle at rest gives: $$E_0 = mc^2$$ Applied to a massive particle, subtracting its rest energy gives $$T = E - E_0 = (\gamma - 1)mc^2$$ Applied to a particle of velocity $v << c$ gives: $$T \approx \frac 1 2 mv^2$$
You can't start with $\frac 1 2 mv^2$ and get to the relativistic equations. But, you can go the other way round. Pretend you learned SR first, and that classical physics is just a special case.

 

[jbergman]

Isn't that just a prejedice of having learned classical physics first? Take the fundamental relation for any particle: $$E^2 = p^2c^2 + m^2c^4$$ Applied to a massless particle gives $$E = pc$$ Applied to a massive particle at rest gives: $$E_0 = mc^2$$ Applied to a massive particle, subtracting its rest energy gives $$T = E - E_0 = (\gamma - 1)mc^2$$ Applied to a particle of velocity $v << c$ gives: $$T \approx \frac 1 2 mv^2$$
You can't start with $\frac 1 2 mv^2$ and get to the relativistic equations. But, you can go the other way round. Pretend you learned SR first, and that classical physics is just a special case.

PeroK, I am willing to accept all those facts (and am familiar with them) if I grokked the derivation of $E^2 = p^2c^2 + m^2c^4$. The wikipedia article's 4-momentum derivation is a bit complex and relies on deriving it from the Lagrangian. As I mentioned above, there seem to be other more physical approaches to deriving these used in earlier papers that I think might be more illuminating.

 

[Sagittarius A-Star]

if I grokked the derivation of $E^2 = p^2c^2 + m^2c^4$.

(1) Four-momentum $\mathbf P = (E/c , p_x, p_y, p_z)$

(2) Minkowski Pseudo-scalar-product $* c^2$:
$$ m^2c^4 := \mathbf P \cdot \mathbf P * c^2 = E^2 - p^2c^2$$

also, with $\vec p = \frac {E} {c^2} \vec v$, from which follows $p^2 = \frac{E^2}{c^4} * v^2$:

(3) $\ \ \ m^2c^4 = E^2 - E^2 * \frac{v^2}{c^2}$

(4) $\ \ \ mc^2 = E \sqrt{1-\frac{v^2}{c^2}} = \begin{cases} 0 & \text{if } v=c \\ E_0 & \text{if } v=0 \\ E_0 & \text{if } v>0 \ \ \ and \ \ \ v<c \ (invariant, E\ depends\ on\ v) \end{cases}$

 

[jbergman]

(1) Four-momentum $\mathbf P = (E/c , p_x, p_y, p_z)$

But how do you show that the time component of P is E/c? That was kind of the point of the whole question. The rest is relatively straightforward after that.

 

[Sagittarius A-Star]

But how do you show that the time component of P is E/c? That was kind of the point of the whole question. The rest is relatively straightforward after that.

• Change of $\vec p$ is force integrated over time.
• Change of $E$ (for example kinetic energy of a massive object) is force integrated over distance in direction of movement. So it is the same, only with spatial dimension instead of time dimension.

If the force is zero, then momentum and energy are constant (=integration constant).

In classical mechanics, mass, momentum and energy were 3 different physical quantities with apparently 3 independent conservation laws. In SR, it's all about the conservation of the four-momentum.

 

[PeroK]

But how do you show that the time component of P is E/c? That was kind of the point of the whole question. The rest is relatively straightforward after that.

That's the new definition of total energy.

It's a conserved quantity. If you take out the rest energy and use a low velocity approximation you get $\frac 1 2 mv^2$.

And, of course, $$\frac{dE}{dt} = \vec F \cdot \vec v$$
What more do you want?

 

[Sagittarius A-Star]

But how do you show that the time component of P is E/c? That was kind of the point of the whole question. The rest is relatively straightforward after that.

Another explanation (for massive particles):

four-momentum $\mathbf P = \frac{E_0}{c^2} * \mathbf U$ (with U = four-velocity, energy is inertial, you don't need to call it "mass")

$\mathbf P = \frac{E_0}{c^2} * \frac {d\mathbf X}{d\tau}$ (with X = four-position)

The following is defined also for photons:

$\mathbf P = \frac{E}{c^2} * \frac {d\mathbf X}{dt}$ (with X = four-position)

$\mathbf P = \frac{E}{c^2} * \frac {d}{dt} (ct, x, y, z) = (E/c, p_x, p_y, p_z)$

 

[jbergman]

• Change of $\vec p$ is force integrated over time.
• Change of $E$ (for example kinetic energy of a massive object) is force integrated over distance in direction of movement. So it is the same, only with spatial dimension instead of time dimension.

If the force is zero, then momentum and energy are constant (=integration constant).

In classical mechanics, mass, momentum and energy were 3 different physical quantities with apparently 3 independent conservation laws. In SR, it's all about the conservation of the four-momentum.

This gave me something to chew on. I need to spend some time digesting this.

 

[Sagittarius A-Star]

This gave me something to chew on. I need to spend some time digesting this.

Example from classical mechanics:

potential energy (in homogenous field)= $\int \vec F \cdot d\vec s = \int (m*\vec g) \, d\vec h = m*g*h + const_1$

kinetic energy = $\int \vec F \cdot d\vec s = \int (m*\vec a) \, d\vec s = \int (m*\frac{d\vec v}{dt}) \, d\vec s = m\int \vec v \cdot d\vec v = \frac{1}{2}mv^2 + const_2$

momentum = $\int \vec F dt = \int (m*\vec a) dt = m*\vec v + const_3$

 

[vanhees71]

You're mixing your theories here. In classical EM, light is a wave, with a wavelength given by the distance between wavefronts. In SR, photons are modeled as massless particles, with the frequency and wavelength being parameters related to the energy-momentum of the photon. The wavelength is in no way the distance between adjacent photons. The number of photons per second is the intensity of the light.

Classical electrodynamics is a relativistic field theory, being unfortunately deformed by many textbooks using non-relativistic approximations for the "matter part".

Photons are particular states (single-photon Fock states) of the quantized electromagnetic field.

Everything that's asked here can be understood within classical electrodynamics. In an arbitrary frame of referece a plane-wave mode of the elektromagnetic field reads
$$\vec{E}=\vec{E}_0 \exp(-\mathrm{i} k \cdot x), \quad \vec{B}=\vec{k} \times \vec{E}, \quad \vec{k} \cdot \vec{E}_0=0.$$
If you only ask for the frequency an arbitrarily moving observer measures, it's given by
$$\omega_{\text{obs}}(\vec{\beta})=u \cdot k = \frac{1}{\sqrt{1-\beta^2}} c k (1-\vec{\beta} \cdot \vec{k}),$$
where $k=|\vec{k}|$ and $\vec{\beta}=\vec{v}/c$. For $\vec{\beta}=0$ you get $\omega=\omega_0=c k$.

Now $\vec{\beta} \cdot \vec{k}=\beta k \cos \vartheta$, where $\vartheta$ is the angle between the direction of wave propagation and the velocity of the observer. For $\vartheta=0$ you have
$$\omega_{\text{obs}}=\sqrt{\frac{1-\beta}{1+\beta}}\omega_0.$$
This is the maximal Doppler red-shift you can get. Thinking of the plane wave as the field emitted from an object that is very far away, the observer moves away from this source and the frequency $\omega_0$ measured in the rest frame of the source is red-shifted to $\omega_{\text{obs}}$.

For $\vartheta=\pi$ the observer moves straight towards the source and thus measures the maximally blue-shifted frequency
$$\omega_{\text{obs}}=\sqrt{\frac{1+\beta}{1-\beta}} \omega_0.$$
For an angle in between you get some red or blue shift within these extreme values.

A completely relativistic effect is that you also get a Doppler red shift for $\vartheta=\pi/2$,
$$\omega_{\text{obs}}=\frac{1}{\sqrt{1-\beta^2}} \omega_0.$$
This is the effect of relativistic time-dilation.

 

[Sagittarius A-Star]

But how do you show that the time component of P is E/c? That was kind of the point of the whole question. The rest is relatively straightforward after that.

Another way to explain it:

(1) Defintion of relativistic 3-momentum (and avoiding the word "relativistic mass"):
$\ \ \ \ \ \ \vec p = \frac {E} {c^2} \vec v$, from which follows $p = \frac{E}{c^2} * v$

(2) Four-momentum $\mathbf P = (p_t , p_x, p_y, p_z)$

(3) Derivative of four-position: $\ \ \ \ \frac{d}{dt} \mathbf X = \frac{d}{dt}(ct, x, y, z) = (c, v_x, v_y, v_z)$

$\ \ \$(3) => An object moves with speed $v= \sqrt{v_x^2+v_y^2+v_z^2}$ in space
$\ \ \ \ \ \$and with velocity $c$ in ct-direction. Therefore:

(4) $\ \ \ p_t = \frac{c}{v} * p$, and with (1):

(5) $\ \ \ p_t = \frac{c}{v} * \frac{E}{c^2} * v = E/c$

(6) Four-momentum $\mathbf P = (E/c , p_x, p_y, p_z)$